# Answer to Question #6241 in Trigonometry for joy

Question #6241

sin(x)^2=2cos(x)+2

Expert's answer

Due to the famous identity cos

So sin

Let's denote cos x=t. We suppose that -1<=t<=1

So we get the equation:

1-t^2=2t+2

t^2+2t+1=0

(t+1)^2=0

t=-1

So cos x=-1

x=pi+2k pi where k - integer

Also we created video where this particular problem is considered. Please take a look!

^{2}x+sin^{2}x=1 therefore sin^{2}x=1-cos^{2}xSo sin

^{2}x=1-cos^{2}x =2cos x+2Let's denote cos x=t. We suppose that -1<=t<=1

So we get the equation:

1-t^2=2t+2

t^2+2t+1=0

(t+1)^2=0

t=-1

So cos x=-1

x=pi+2k pi where k - integer

Also we created video where this particular problem is considered. Please take a look!

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